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Combination formula on dice and bit errors

  1. Mar 29, 2015 #1
    When the learning combination formula the formula nCr is usually derived using a deck of cards. There are n=52 different card. If you choose r number of cards, each is different so you basically start with n!/(n-r)! , then because the order is unimportant multiply by 1/r!. This explanation is pretty straight forward. What I'm having trouble wrapping my head around is how the formula is applied to bit errors where there seems to be repeats and order does matter. For example if I want to find the probability of having one bit error in a four bit word I start by finding the number of possible ways there are of having single bit error in the word.
    Where 'o' represents a correct bit and 'E' represents an error, here are the possible ways to have words with a single bit error:
    oooE ooEo ooEo Eooo
    Counting the ways and using the formula 4C1 i find there are 4 combinations. This works also for 2,3, and 4 errors, but I don't understand why the combination formula applies, as it seems that order does matter, and there does seem to be repeats. Should I look at the situation as each bit error position is unique like each card in a deck is unique, so that when I "choose" r bit errors I'm actually choosing the bit position to be in error.

    Also for the situation with something like dice. It seems I should be able to extend the same concept as above. Lets say I have three, three sided dice. If I want to calculate the odds of rolling a single '2'. Should I use the combination formula to find how many ways there are of rolling a single '2'? If I tabulate the possible ways of rolling the dice and count the ways having a single '2' in the outcome I count 12 ways (underlined). But if I use 3C1 I of course get 3 as the number of combinations of outcomes that produce a single 2. It seems obvious that the probability of rolling a single '2' is 12/27 and not 3/27, but what method should I use find the possible ways of getting a single '2' in this situation. It seems that I'm just extending the concept of how I used nCr to find the possible ways of having a single bit in error.

    \begin{matrix} 111& \underline{112} & 113 \\ \underline{121}& 122& \underline{123}\\ 131& \underline{132} & 133 \\ \underline{211}& 212 & \underline{213} \\ 221& 222& 223\\ \underline{231}& 232 & \underline{233} \\ 311& \underline{312} & 313 \\ \underline{321}& 322& \underline{323}\\ 331& \underline{332} & 333 \\ & & \end{matrix}
     
    Last edited: Mar 29, 2015
  2. jcsd
  3. Mar 29, 2015 #2

    mfb

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    Staff: Mentor

    "Bit order" does not correspond to "order of drawing cards".
    "Bit order" is equivalent to "all 52 cards are different" (imagine 52 bits, each bit is like a card) and "the order of cards drawn" does not matter corresponds to "if we pick two bits at random to have an error, it does not matter which one we pick first".
     
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